Find the area of the part of the surface $z=x^2+2y$ that lies above the triangle with vertices $(0,0)$, $(1,0)$, $(1,2)$.
Not really sure where to start with this question. So far i have my bounds for $x$ from $0-1$ and for $y$ $0-2$. I have my double integral over $D$ set up but after this I don't really know what to do.
I have my $x^2 +2y$ under a square root with them being squared. But when I try to integrate that it becomes a mess and just does not seem like it is right.
Please help!!!!!!!!
It sounds like you have made a mess.
$\iint \|dS\|\ dx\ dy$
$dS = (-\frac {\partial z}{\partial x},-\frac {\partial z}{\partial y},1) = (-x,-2, 1)\\ \|dS\| = \sqrt {x^2 + 2}$
your region is bound by the lines, $y = 0, y = 2x,x = 1$
$\int_0^1\int_0^{2x} \sqrt {x^2 + 2}\ dy\ dx$